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Lecture 14: Query Optimization. This Lecture. Query rewriting Cost estimation We have learned how atomic operations are implemented and their cost We’ll complete the picture by estimating the cost query plans Including computing the size of the output (reduction factors) - PowerPoint PPT Presentation
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This Lecture
• Query rewriting• Cost estimation
– We have learned how atomic operations are implemented and their cost
– We’ll complete the picture by estimating the cost query plans
– Including computing the size of the output (reduction factors)
• Join ordering - optimization
Query Optimization Process(simplified a bit)
• Parse the SQL query into a logical tree:– identify distinct blocks (corresponding to nested sub-
queries or views). • Query rewrite phase: SQL-to-SQL transforms
– apply algebraic transformations to yield a cheaper plan.– Merge blocks and move predicates between blocks.
• Optimize each block: join ordering.• Complete the optimization: select scheduling
(pipelining strategy).
Operations (revisited)
• Scan ([index], table, predicate):– Either index scan or table scan.– Try to push down sargable predicates.
• Selection (filter)• Projection (always need to go to the data?)• Joins: nested loop (indexed), sort-merge,
hash, outer join.• Grouping and aggregation (usually the last).
Algebraic Laws
• Commutative and Associative Laws– R U S = S U R, R U (S U T) = (R U S) U T– R ∩ S = S ∩ R, R ∩ (S ∩ T) = (R ∩ S) ∩ T– R ⋈ S = S ⋈ R, R ⋈ (S ⋈ T) = (R ⋈ S) ⋈
T• Distributive Laws
– R ⋈ (S U T) = (R ⋈ S) U (R ⋈ T)
Algebraic Laws
• Laws involving selection:– s C AND C’(R) = s C(s C’(R)) = s C(R) ∩ s C’(R)– s C OR C’(R) = s C(R) U s C’(R)– s C (R ⋈ S) = s C (R) ⋈ S
• When C involves only attributes of R– s C (R – S) = s C (R) – S– s C (R U S) = s C (R) U s C (S)– s C (R ∩ S) = s C (R) ∩ S
Algebraic Laws
• Example: R(A, B, C, D), S(E, F, G)– s F=3 (R ⋈ S) = ?– s A=5 AND G=9 (R ⋈ S) = ?
D=E
D=E
Algebraic Laws
• Laws involving projections– PM(R ⋈ S) = PN(PP(R) ⋈ PQ(S))
• Where N, P, Q are appropriate subsets of attributes of M
– PM(PN(R)) = PM∩N (R)• Example: R(A,B,C,D), S(E, F, G)
– PA,B,G(R ⋈ S) = P ? (P?(R) ⋈ P?(S)) D=E D=E
Query Rewrites: Sub-queries
SELECT Emp.NameFROM EmpWHERE Emp.Age < 30 AND
Emp.Dept# IN(SELECT Dept.Dept#FROM DeptWHERE Dept.Loc = “Seattle”AND Emp.Emp#=Dept.Mgr)
The Un-Nested Query
SELECT Emp.NameFROM Emp, DeptWHERE Emp.Age < 30 AND Emp.Dept#=Dept.Dept# AND Dept.Loc = “Seattle” AND Emp.Emp#=Dept.Mgr
Converting Nested Queries
Select distinct x.name, x.makerFrom product xWhere x.color= “blue” AND x.price >= ALL (Select y.price From product y Where x.maker = y.maker AND y.color=“blue”)
Converting Nested Queries
Select distinct x.name, x.makerFrom product xWhere x.color= “blue” AND x.price < SOME (Select y.price From product y Where x.maker = y.maker AND y.color=“blue”)
Let’s compute the complement first:
Converting Nested Queries
Select distinct x.name, x.makerFrom product x, product yWhere x.color= “blue” AND x.maker = y.maker AND y.color=“blue” AND x.price < y.price
This one becomes a SFW query:
This returns exactly the products we DON’T want, so…
Converting Nested Queries
(Select x.name, x.maker From product x Where x.color = “blue”)
EXCEPT
(Select x.name, x.maker From product x, product y Where x.color= “blue” AND x.maker = y.maker AND y.color=“blue” AND x.price < y.price)
Semi-Joins, Magic Sets
• You can’t always un-nest sub-queries (it’s tricky).• But you can often use a semi-join to reduce the
computation cost of the inner query.• A magic set is a superset of the possible bindings
in the result of the sub-query.• Also called “sideways information passing”.• Great idea; reinvented every few years on a
regular basis.
Rewrites: Magic SetsCreate View DepAvgSal AS (Select E.did, Avg(E.sal) as avgsal From Emp E Group By E.did)
Select E.eid, E.salFrom Emp E, Dept D, DepAvgSal VWhere E.did=D.did AND D.did=V.did And E.age < 30 and D.budget > 100k And E.sal > V.avgsal
Rewrites: SIPsSelect E.eid, E.salFrom Emp E, Dept D, DepAvgSal VWhere E.did=D.did AND D.did=V.did And E.age < 30 and D.budget > 100k And E.sal > V.avgsal• DepAvgsal needs to be evaluated only for
departments where V.did IN Select E.did From Emp E, Dept D Where E.did=D.did And E.age < 30 and D.budget > 100K
Supporting Views1. Create View PartialResult as (Select E.eid, E.sal, E.did From Emp E, Dept D Where E.did=D.did And E.age < 30 and D.budget > 100K)2. Create View Filter AS Select DISTINCT P.did FROM PartialResult P.3. Create View LimitedAvgSal as (Select F.did, Avg(E.Sal) as avgSal From Emp E, Filter F Where E.did=F.did Group By F.did)
And Finally…
Transformed query:
Select P.eid, P.sal From PartialResult P, LimitedAvgSal V Where P.did=V.did And P.sal > V.avgsal
Rewrites: Group By and Join• Schema:
– Product (pid, unitprice,…)– Sales(tid, date, store, pid, units)
• Trees:
Join
groupBy(pid)Sum(units)
Scan(Sales)Filter(date in Q2,2013)
ProductsFilter (price>100)
Join
groupBy(pid)Sum(units)
Scan(Sales)Filter(date in Q2,2013)
ProductsFilter (price>100)
Schema for Some Examples
• Reserves:– Each tuple is 40 bytes long, 100 tuples per page, 1000
pages• Sailors:
– Each tuple is 50 bytes long, 80 tuples per page, 500 pages
Sailors (sid: integer, sname: string, rating: integer, age: real)Reserves (sid: integer, bid: integer, day: date, rname: string)
Query Rewriting: Predicate Pushdown
Reserves Sailors
sid=sid
σbid=100 AND rating >5
Psname
sid=sid
Psname
Reserves Sailors
σbid=100 σrating > 5
(Scan; write to temp T1)
The earlier we process selections, the less tuples we need to manipulatehigher up in the tree.Disadvantages?
(Scan; write to temp T2)
Query Rewrites: Predicate Pushdown (through grouping)
Select bid, Max(age)From Reserves R, Sailors SWhere R.sid=S.sid GroupBy bidHaving Max(age) > 40
Select bid, Max(age)From Reserves R, Sailors SWhere R.sid=S.sid and S.age > 40GroupBy bid
• For each boat, find the maximal age of sailors who’ve reserved it.• Advantage: the size of the join will be smaller.• Requires transformation rules specific to the grouping/aggregation operators.• Will it work work if we replace Max by Min?
Query Rewrite:Predicate Movearound
Sailing dates: when did the youngest of each sailor level rent boats?
Create View V1 ASSelect rating, Min(age)From Sailors SWhere S.age < 20Group By rating
Create View V2 ASSelect sid, rating, age, dateFrom Sailors S, Reserves RWhere R.sid=S.sid
Select sid, dateFrom V1, V2Where V1.rating = V2.rating and V1.age = V2.age
Query Rewrite: Predicate Movearound
Create View V1 ASSelect rating, Min(age)From Sailors SWhere S.age < 20Group By rating
Create View V2 ASSelect sid, rating, age, dateFrom Sailors S, Reserves RWhere R.sid=S.sid
Select sid, dateFrom V1, V2Where V1.rating = V2.rating and V1.age = V2.age, age < 20
Sailing dates: when did the youngest of each sailor level rent boats?
First, move predicates up the tree.
Query Rewrite: Predicate Movearound
Create View V1 ASSelect rating, Min(age)From Sailors SWhere S.age < 20Group By rating
Create View V2 ASSelect sid, rating, age, dateFrom Sailors S, Reserves RWhere R.sid=S.sid, and S.age < 20.
Select sid, dateFrom V1, V2Where V1.rating = V2.rating and V1.age = V2.age, and age < 20
Sailing dates: when did the youngest of each sailor level rent boats?
First, move predicates up the tree.
Then, move themdown.
Query Rewrite Summary• The optimizer can use any semantically correct
rule to transform one query to another.• Rules are used for:
– moving constraints between blocks (because each will be optimized separately)
– Un-nesting blocks• In a few minutes of thought, you’ll come up with
your own rewrite. Some query, somewhere, will benefit from it.
• Theorems?
Cost Estimation• For each plan considered, must estimate cost:
– Must estimate cost of each operation in plan tree.• Depends on input cardinalities.
– Must estimate size of result for each operation in tree!• Use information about the input relations.• For selections and joins, assume independence of predicates.
• We’ll discuss the System R cost estimation approach.– Very inexact, but works ok in practice.– More sophisticated techniques known now.
Statistics and Catalogs• Need information about the relations and indexes
involved. Catalogs typically contain at least:– # tuples (T) and # pages (B) for each relation.– # distinct key values (K) and #pages for each index.– Index height, low/high key values (Low/High) for each tree
index.– Alternatively V(R, F) – #distinct values in field F of relation R
• Catalogs updated periodically.– Updating whenever data changes is too expensive; lots of
approximation anyway, so slight inconsistency ok.• More detailed information (e.g., histograms of the values
in some field) are sometimes stored.
Size Estimation and Reduction Factors
• Consider a query block:• Maximum # tuples in result is the product of the
cardinalities of relations in the FROM clause.• Reduction factor (RF) associated with each term reflects
the impact of the term in reducing result size. Result cardinality = Max # tuples · product of all RF’s.– Implicit assumption that terms are independent!– Term col=value has RF 1/K(I), given index I on col– Term col1=col2 has RF 1/MAX(K(I1), K(I2))– Term col>value has RF (High(I)-value)/(High(I)-Low(I))
SELECT attribute listFROM relation listWHERE term1 AND ... AND termk
Histograms
• Key to obtaining good cost and size estimates.
• Come in several flavors:– Equi-depth– Equi-width
• Which is better?• Compressed histograms: special treatment
of frequent values.
Histograms
• Statistics on data maintained by the RDBMS
• Makes size estimation much more accurate (hence, cost estimations are more accurate)
• V(R,F) – number of distinct values in field F of relation R
Histograms
Employee(ssn, name, salary, phone)• Maintain a histogram on salary:
• T(Employee) = 25000, but now we know the distribution
Salary: 0..20k 20k..40k 40k..60k 60k..80k 80k..100k > 100k
Tuples 200 800 5000 12000 6500 500
Histograms
Ranks(rankName, salary)• Estimate the size of Employee ⋈ Ranks
Employee 0..20k 20k..40k 40k..60k 60k..80k 80k..100k > 100k
200 800 5000 12000 6500 500
Ranks 0..20k 20k..40k 40k..60k 60k..80k 80k..100k > 100k
8 20 40 80 100 2
Salary
Histograms
• Assume:– V(Employee, Salary) = 200– V(Ranks, Salary) = 250
• Then T(Employee ⋈ Ranks) == (Si=1,6 Ti T’i)/ 250= (200x8 + 800x20 + 5000x40 + 12000x80 + 6500x100 +
500x2)/250= ….
Salary
Schema for Some Examples
• Reserves:– Each tuple is 40 bytes long, 100 tuples per page, 1000
pages• Sailors:
– Each tuple is 50 bytes long, 80 tuples per page, 500 pages
Sailors (sid: integer, sname: string, rating: integer, age: real)Reserves (sid: integer, bid: integer, day: date, rname: string)
Pipelining• Assume that we want to find all the boats ever reserved by
sailors with rank 8.SELECT R.bidFROM Sailors S, Reserves RWHERE S.rating=8 AND S.sid=R.sid• Consider a plan that performs selection, hash join, and
projection in this order• Writing the result of each operation to the disk is redundant!• In the best case, there is enough memory to perform all the
operations simultaneously in memory
Pipelining
Sailors Reserves
σrating=8
h2Пbid
Output
2
1
N
h
2
1
hN
SelectionHash join step 1 for Sailors
Hash join step 1 for Reserves Hash join step 2 Projection
Pipelining• We ignored output cost for single operations• For operations that get their input in a pipeline, we
reduce the first read cost– E.g., selection is for free!– In the previous example – we “saved” the full read in the first
stage of hash join for sailor, and the full read for the projection
• Requires sufficient memory– To perform succeeding operations in parallel– We will usually assume that this is the case
Plan Optimimization• Task: create a query execution plan• Key idea: perform an estimation of the possible
query plan costs and choose the cheapest one• The query plan operations are carried out in some
order, where the results of one operation are pipelined
Example• If we have an Index on rating:
– (1/K(I)) · T(R) = (1/10) · 40000 tuples retrieved.– Clustered index: (1/K(I)) · (B(I)+B(R)) = search cost + (1/10) ·
(500) pages are read (= 51.2-54). – Unclustered index: (1/K(I)) · (B(I)+T(R)) =
search cost + (1/10) · (50+40000) pages are read. • Doing a file scan: we retrieve all the pages (500).
SELECT S.sidFROM Sailors SWHERE S.rating=8
Determining Join Ordering
• R1 R2 …. Rn• Join tree:
• A join tree represents a plan. An optimizer needs to inspect many (all ?) join trees
R3 R1 R2 R4
Problem
• Given: a query R1 ⋈ R2 ⋈ … ⋈ Rn
• Assume we have a function cost() that gives us the cost of every join tree
• Find the best join tree for the query
Dynamic Programming
• Idea: for each subset of {R1, …, Rn}, compute the best plan for that subset
• In increasing order of set cardinality:– Step 1: for {R1}, {R2}, …, {Rn}– Step 2: for {R1,R2}, {R1,R3}, …, {Rn-1, Rn}– …– Step n: for {R1, …, Rn}
• A subset of {R1, …, Rn} is also called a subquery
Dynamic Programming
• For each subquery Q {R⊆ 1, …, Rn} compute the following:– Size(Q)– A best plan for Q: Plan(Q)– The cost of that plan: Cost(Q)
Dynamic Programming
• Step 1: For each {Ri} do:– Size({Ri}) = B(Ri)– Plan({Ri}) = Ri
– Cost({Ri}) = (cost of scanning Ri)
Dynamic Programming
• Step i: For each Q {R⊆ 1, …, Rn} of cardinality i do:– Compute Size(Q) – For every pair of subqueries Q’, Q’’
s.t. Q = Q’ U Q’’compute cost(Plan(Q’) ⋈ Plan(Q’’))
– Cost(Q) = the smallest such cost– Plan(Q) = the corresponding plan
Dynamic Programming• Summary: computes optimal plans for subqueries:
– Step 1: {R1}, {R2}, …, {Rn}– Step 2: {R1, R2}, {R1, R3}, …, {Rn-1, Rn}– …– Step n: {R1, …, Rn}
• We used naïve size/cost estimations• In practice:
– more realistic size/cost estimations (next)– heuristics for Reducing the Search Space
• Restrict to left linear trees• Restrict to trees “without cartesian product”
– need more than just one plan for each subquery:• “interesting orders”